CN102564461A - Method for calibrating optical strapdown inertial navigation system based on two-axis turntable - Google Patents

Abstract

The utility model relates to a method for calibrating an optical strapdown inertial navigation system based on a two-axis turntable. The method comprises the following six steps: firstly, selecting an error model; secondly, mounting three axles of the optical strapdown inertial navigation system to the two-axis turntable according to the east-north-up direction, and then collecting data after the whole system is powered on and preheated; thirdly, collecting data and storing the data to a calibration matrix according to the twelve positions of east-north-up, west-south-up, north-east-down, south-west-down, up-south-east, up-north-west, down-north-east, down-south-west, south-up-west, north-up-east, south-down-east and north-down-west; fourthly, rotating around the z-axis, the y-axis or the x-axis for 3 minutes at omega of plus or minus 2 degrees per seconds, plus or minus 10 degrees per seconds, plus or minus 30 degrees per seconds, plus or minus 60 degrees per seconds or plus or minus 100 degrees per seconds according to the two positions of east-north-up and north-west-up, and equalizing the output data and storing the average value in the calibration matrix; fifthly, collecting data for 15 minutes at each position of the four positions of east-north-up, west-north-up, north-east-down and south-west-down, and then equalizing the output data and storing the average value in the calibration matrix; and sixthly, using the least square method for calculating the calibration parameters of an accelerometer, and using an analysis method for calculating the calibration parameters of a gyroscope.

Description

A kind of scaling method of the optics SINS based on double axle table

(1) technical field

The present invention relates to a kind of scaling method of the optics SINS based on double axle table, belong to the Aeronautics and Astronautics navigation instrument and demarcate and the detection technique field.

(2) background technology

Calibration technique is a kind of Error Compensation Technology in essence.The so-called demarcation is the model equation of setting up inertia type instrument and inertia system, utilizes special testing apparatus and software algorithm, calibrates the error term of instrument and system, and recompenses, and improves the actual service precision of instrument and system.At present scaling method need utilize turntable to carry out pre-set demarcation path, and through position measurement test, speed measurement test, the zero-bit that calibrates optical gyroscope is value, scale factor, alignment error partially; The zero-bit of quartz flexible accelerometer is value, scale factor, alignment error partially.

The calibration technique of optics SINS is at first set up the input about error term as a kind of Error Compensation Technology, and optics SINS error model is the correlated error item mathematics model of setting up through to the system performance analysis.After error model is set up, adopt different scaling methods and special-purpose testing apparatus, pick out the error term in the error model, wherein testing apparatus is used to provide relatively accurate input reference, uses turntable as special test equipment usually.After picking out each error term, utilize error model to compensate in the inertial navigation output data, obtain the higher data of precision.

Optics SINS error model.The optical gyroscope error model: optical gyroscope comprises optical fibre gyro and laser gyro, because the design feature between the two, the error model difference is little, and most laser, optical fibre gyro error models of setting up are identical.Usually choose scale factor, misalignment, zero partially as error term.The accelerometer error model: the modeling for quartz flexible accelerometer has difference to set up mode, and the key distinction is whether to contain the quadratic term error of specific force, but in general this be in a small amount, and is little to the calibration result influence.Identical with Gyro Calibration, choose scale factor, misalignment, zero usually partially as error term.

Turntable is the visual plant of demarcating, and turntable is the device that accurate position, rotating speed are provided for measured equipment.We utilize turntable is in order to obtain inertial navigation system input information accurately, and utilizes diverse location, rotating speed that different error term is separated, and reaches the purpose of accurately obtaining calibrating parameters.Turntable has following vital role to the inertial navigation system test calibration:

1) accurately characterizes the directivity of carrier at inertial space medium speed and acceleration.

2) inertial navigation system input information accurately is provided continuously, for inertial navigation system error compensation provides reference.

Present scaling method designs based on three-axle table mostly, and Cao Ningsheng, Chen Beiou, Deng Zhihong have designed the layout of speed experiment and multiposition static test respectively, demarcate gyro and accelerometer parameter.In order to shorten the nominal time, some scholar combines rate test and multiposition test, has designed the position and has intersected with rotation and experimentize, and demarcates gyro and accelerometer parameter simultaneously.Like the six positions rotation scaling method of Liu Baiqi, " the positive and negative rotating speed in six orientation " method of Li Jianli, " six positions of the dynamically rolling " quick calibrating method of Sun Hongwei and seven positions of Yan Gongmin are rotated the demarcation scheme continuously.

Three-axle table possesses comprehensive spatial rotation and positioning function, promptly can realize the attitude angle motion that carrier is all.But three-axle table exists and to cost an arm and a leg, and problems such as bulky, installation and maintenance difficulty restrict its application on inertial navigation system is demarcated.Double axle table lacks one degree of freedom than three-axle table, and its location is not enough with rotating function.But use the turntable angle from demarcating, common scaling method can not utilize the whole location and the rotating function of three-axle table, and many times double axle table can obtain the calibration result of equal accuracy equally.And the double axle table of equal accuracy is low, low in energy consumption than the three-axle table price, volume is little.

Utilize the scaling method of three-axle table to be difficult to overcome the influence that turntable self error is brought, because three-axle table has three degree of freedom, total 24 errors that comprise the error of perpendicularity, intersect degree error, turn error.Wherein the error of perpendicularity, turn error are difficult to eliminate in above-mentioned scaling method, and then influence the stated accuracy of inertial navigation system.

The shortcoming of prior art:

1) is difficult to eliminate the influence of turntable self error based on the three-axle table scaling method, and then influences stated accuracy;

2) three-axle table exist cost an arm and a leg, shortcomings such as volume is big, installation and maintenance difficulty;

3), also do not take into account the scaling method of its rotating function and accuracy requirement at present based on double axle table design from existing document.

(3) summary of the invention

The invention provides a kind of scaling method of the optics SINS based on double axle table, this calibration technique can effectively be eliminated the influence of turntable alignment error, obtains the calibration result of degree of precision.

The scaling method of a kind of optics SINS based on double axle table of the present invention, its concrete cloth is following suddenly:

Step 1: the choosing of error model;

The accelerometer error model:

$\left(\begin{array}{c}{f}_{\mathrm{ax}}\\ f_{\mathrm{ay}}\\ f_{\mathrm{az}}\end{array}\right)=\left(\begin{array}{ccc}{k}_{\mathrm{axx}}& k_{\mathrm{axy}}& k_{\mathrm{axz}}\\ k_{\mathrm{ayx}}& k_{\mathrm{ayy}}& k_{\mathrm{ayz}}\\ k_{\mathrm{azx}}& k_{\mathrm{azy}}& k_{\mathrm{azz}}\end{array}\right)\·\left(\begin{array}{c}{A}_x\\ A_y\\ A_z\end{array}\right)+\left(\begin{array}{c}{B}_{\mathrm{ax}}\\ B_{\mathrm{ay}}\\ B_{\mathrm{az}}\end{array}\right)...\left(1\right)$

Gyroscope error model:

$\left(\begin{array}{c}{f}_{\mathrm{gx}}\\ f_{\mathrm{gy}}\\ f_{\mathrm{gz}}\end{array}\right)=\left(\begin{array}{ccc}{k}_{\mathrm{gxx}}& k_{\mathrm{gxy}}& k_{\mathrm{gxz}}\\ k_{\mathrm{gyx}}& k_{\mathrm{gyy}}& k_{\mathrm{gyz}}\\ k_{\mathrm{gzx}}& k_{\mathrm{gzy}}& k_{\mathrm{gzz}}\end{array}\right)\·\left(\begin{array}{c}{\mathrm{\ω}}_x\\ {\mathrm{\ω}}_y\\ {\mathrm{\ω}}_z\end{array}\right)+\left(\begin{array}{c}{B}_{\mathrm{gx}}\\ B_{\mathrm{gy}}\\ B_{\mathrm{gz}}\end{array}\right)...\left(2\right)$

F wherein _Ai, f _GiBe respectively i axis accelerometer, gyro output, A _i, ω _iBe respectively i axis accelerometer, gyro input, B _Ai, B _GiBe respectively i axis accelerometer, gyro zero partially, k _Aii, k _GiiBe respectively input i axis accelerometer, the projection of gyro unit's input on output i axle, k _Aij, k _GijBe respectively projection on output i axle of input j axis accelerometer, the input of gyro unit (i, j=x, y, z).

Step 2: direction is installed on the double axle table towards the sky, northeast with three of inertial navigation systems, and turntable, inertial navigation system, collecting computer connect and finish, and the energising preheating got into data acquisition after 30 minutes;

Step 3: according to 12 positions shown in Figure 1 be sky, northeast, southwest sky, east northeast ground, Nan Xidi, day east southeast, sky northwest (NW), east northeast, Di Nanxi, Nan Tian west, Bei Tiandong, Nan Didong, western 12 positions, backlands; Each position is static gathered three minutes; And after making even data all, store into and demarcate in the matrix;

Step 4: according to sky, northeast, two positions shown in Figure 2, position, northwest (NW) sky, be ± 2 °/s around z, y, x with ω respectively, ± 10 °/s, ± 30 °/s, ± 60 °/s, ± 100 °/s rotated three minutes, stored into after making even output data all and demarcated in the matrix;

Step 5: according to four positions shown in Figure 3 is sky, northeast, southwest sky, east northeast ground, southern western destination location, and static collection 15 minutes on each position is respectively got output data average and stored in the demarcation matrix;

Step 6: utilize least square method to resolve the accelerometer calibrating parameters, utilize analytic calculation Gyro Calibration parameter;

1) utilize least square method to resolve the accelerometer calibrating parameters:

Utilize the measurement result of 12 positions in the step 3, wherein

expression x axle is towards the mean value of all position j axle outputs in sky;

expression x axle is towards the mean value of all position j axle outputs on ground;

expression y axle is towards the mean value of all position j axle outputs in sky;

expression y axle is towards the mean value of all position j axle outputs on ground;

expression z axle is towards the mean value of all position j axle outputs in sky;

expression z axle is towards the mean value of all position j axle outputs on ground.

$B_{\mathrm{ax}}=({\stackrel{\‾}{f}}_{\mathrm{ax}}^{x+}+{\stackrel{\‾}{f}}_{\mathrm{ax}}^{x-}+{\stackrel{\‾}{f}}_{\mathrm{ax}}^{y+}+{\stackrel{\‾}{f}}_{\mathrm{ax}}^{y-}+{\stackrel{\‾}{f}}_{\mathrm{ax}}^{z+}+{\stackrel{\‾}{f}}_{\mathrm{ax}}^{z-})/6...\left(3\right)$

$k_{\mathrm{axx}}=({\stackrel{\‾}{f}}_{\mathrm{ax}}^{x+}-{\stackrel{\‾}{f}}_{\mathrm{ax}}^{x-})/2g...\left(4\right)$

$k_{\mathrm{axy}}=({\stackrel{\‾}{f}}_{\mathrm{ax}}^{y+}-{\stackrel{\‾}{f}}_{\mathrm{ax}}^{y-})/2g...\left(5\right)$

$k_{\mathrm{axz}}=({\stackrel{\‾}{f}}_{\mathrm{ax}}^{z+}-{\stackrel{\‾}{f}}_{\mathrm{ax}}^{z-})/2g...\left(6\right)$

$B_{\mathrm{ay}}=({\stackrel{\‾}{f}}_{\mathrm{ay}}^{x+}+{\stackrel{\‾}{f}}_{\mathrm{ay}}^{x-}+{\stackrel{\‾}{f}}_{\mathrm{ay}}^{y+}+{\stackrel{\‾}{f}}_{\mathrm{ay}}^{y-}+{\stackrel{\‾}{f}}_{\mathrm{ay}}^{z+}+{\stackrel{\‾}{f}}_{\mathrm{ay}}^{z-})/6...\left(7\right)$

$k_{\mathrm{ayx}}=({\stackrel{\‾}{f}}_{\mathrm{ay}}^{x+}-{\stackrel{\‾}{f}}_{\mathrm{ay}}^{x-})/2g...\left(8\right)$

$k_{\mathrm{ayy}}=({\stackrel{\‾}{f}}_{\mathrm{ay}}^{y+}-{\stackrel{\‾}{f}}_{\mathrm{ay}}^{y-})/2g...\left(9\right)$

$k_{\mathrm{ayz}}=({\stackrel{\‾}{f}}_{\mathrm{ay}}^{z+}-{\stackrel{\‾}{f}}_{\mathrm{ay}}^{z-})/2g...\left(10\right)$

$B_{\mathrm{az}}=({\stackrel{\‾}{f}}_{\mathrm{az}}^{x+}+{\stackrel{\‾}{f}}_{\mathrm{az}}^{x-}+{\stackrel{\‾}{f}}_{\mathrm{az}}^{y+}+{\stackrel{\‾}{f}}_{\mathrm{az}}^{y-}+{\stackrel{\‾}{f}}_{\mathrm{az}}^{z+}+{\stackrel{\‾}{f}}_{\mathrm{az}}^{z-})/6...\left(11\right)$

$k_{\mathrm{azx}}=({\stackrel{\‾}{f}}_{\mathrm{az}}^{x+}-{\stackrel{\‾}{f}}_{\mathrm{az}}^{x-})/2g...\left(12\right)$

$k_{\mathrm{azy}}=({\stackrel{\‾}{f}}_{\mathrm{az}}^{y+}-{\stackrel{\‾}{f}}_{\mathrm{az}}^{y-})/2g...\left(13\right)$

$k_{\mathrm{azz}}=({\stackrel{\‾}{f}}_{\mathrm{az}}^{z+}-{\stackrel{\‾}{f}}_{\mathrm{az}}^{z-})/2g...\left(14\right)$

2) utilize analytical method to resolve gyro constant multiplier and misalignment:

Utilize three rotating measurement results in the step 4, wherein

For (j=x, y is z) with angular speed ± ω around j _kBe slewing rate, in rotation n time-of-week, (z) axle is respectively at angular speed ω for i=x, y for i _k,-ω _kMean value poor of output, ω ₁, ω ₂, ω ₃, ω ₄, ω ₅Be respectively 2 °/s, 10 °/s, 30 °/s, 60 °/s, 100 °/s.

$k_{\mathrm{gx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="F\; gxx"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxx}}^{}}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="F\; gxx"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxx}}^{}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="3"\; label="F\; gxx"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxx}}^{}}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="4"\; label="F\; gxx"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxx}}^{}}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="5"\; label="F\; gxx"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxx}}^{}}}{{2\mathrm{\&omega;}}_5})/5...\left(15\right)$

$k_{\mathrm{gy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(16\right)$

$k_{\mathrm{gz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="3"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="4"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="5"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_5})/5...\left(17\right)$

$E_{\mathrm{gxy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="1"\; label="gxy"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">gxy</figure-callout>}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="2"\; label="gxy"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">gxy</figure-callout>}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="3"\; label="gxy"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">gxy</figure-callout>}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="4"\; label="gxy"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">gxy</figure-callout>}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(18\right)$

$E_{\mathrm{gxz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="F\; gxz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxz}}^{}}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="F\; gxz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxz}}^{}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="3"\; label="F\; gxz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxz}}^{}}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="4"\; label="F\; gxz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxz}}^{}}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="5"\; label="F\; gxz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gxz}}^{}}}{{2\mathrm{\&omega;}}_5})/5...\left(19\right)$

$E_{\mathrm{gyx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="1"\; label="gyx"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">gyx</figure-callout>}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="2"\; label="gyx"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">gyx</figure-callout>}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{<figure-callout\; id="5"\; label="gyx"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">gyx</figure-callout>}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(20\right)$

$E_{\mathrm{gyz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="F\; gyz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gyz}}^{}}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="F\; gyz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gyz}}^{}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="3"\; label="F\; gyz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gyz}}^{}}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="4"\; label="F\; gyz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gyz}}^{}}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="5"\; label="F\; gyz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gyz}}^{}}}{{2\mathrm{\&omega;}}_5})/5...\left(21\right)$

$E_{\mathrm{gzx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(22\right)$

$E_{\mathrm{gzy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(23\right)$

3) utilize analytical method to resolve gyro zero partially

B _gx＝(f _1x+f _2x+f _3x+f _4x)/4.............................................(24)

B _gy＝(f _1y+f _2y+f _3y+f _4y)/4.............................................(25)

B _gz＝(f _1z+f _2z+f _3z+f _4z)/4.............................................(26)

Utilize measurement result in the step 5, f _IjExpression i the position output of j axle gyro (i=1,2,3,4; J=x, y, z).

The invention has the advantages that:

1) can effectively eliminate turntable the influence of lateral error to calibration result is installed.The influence of the horizontal alignment error of turntable to the accelerometer calibration result can effectively be reduced in 12 positions, when lateral error be 3.43 ' the constant multiplier error brought is 1 * 10 ^-6(1ppm).When lateral error is 4.05 °, true misalignment is 1000 " time, the misalignment error of generation is 5 ".So this scaling method can effectively suppress the influence of lateral error;

2) can eliminate of the influence of turntable installation position error to calibration result;

3) make full use of the rotation and the positioning function of double axle table, can reach and demarcate effect preferably;

4) the calibration facility cost is low, and required installing space is little, and is simple to operate.

(4) description of drawings

Fig. 1 is 12 position accelerometer calibration position layouts.

Fig. 2 is three rotating gyro constant multipliers, the layout of misalignment calibration position.

Fig. 3 is gyro zero calibration position layout partially.

Fig. 4 is for demarcating synoptic diagram.

Fig. 5 is a FB(flow block) of the present invention.

(5) embodiment

Do further detailed description in the face of the present invention down.

See Fig. 5, the scaling method of a kind of optics SINS based on double axle table of the present invention, the concrete cloth of this method is following suddenly:

Step 1: choose error model

Optics SINS error model is chosen and will be confirmed according to the characteristic of inertia device; Common optics SINS adopts optical gyroscope (laser gyro, optical fibre gyro), quartz flexible accelerometer as inertia device; A measured repeatedly correlated error item magnitude is less, generally only chooses the zero degree item, once as error term.

The accelerometer error model:

$\left(\begin{array}{c}{f}_{\mathrm{ax}}\\ f_{\mathrm{ay}}\\ f_{\mathrm{az}}\end{array}\right)=\left(\begin{array}{ccc}{k}_{\mathrm{axx}}& k_{\mathrm{axy}}& k_{\mathrm{axz}}\\ k_{\mathrm{ayx}}& k_{\mathrm{ayy}}& k_{\mathrm{ayz}}\\ k_{\mathrm{azx}}& k_{\mathrm{azy}}& k_{\mathrm{azz}}\end{array}\right)\&CenterDot;\left(\begin{array}{c}{A}_x\\ A_y\\ A_z\end{array}\right)+\left(\begin{array}{c}{B}_{\mathrm{ax}}\\ B_{\mathrm{ay}}\\ B_{\mathrm{az}}\end{array}\right)...\left(27\right)$

Gyroscope error model:

$\left(\begin{array}{c}{f}_{\mathrm{gx}}\\ f_{\mathrm{gy}}\\ f_{\mathrm{gz}}\end{array}\right)=\left(\begin{array}{ccc}{k}_{\mathrm{gxx}}& k_{\mathrm{gxy}}& k_{\mathrm{gxz}}\\ k_{\mathrm{gyx}}& k_{\mathrm{gyy}}& k_{\mathrm{gyz}}\\ k_{\mathrm{gzx}}& k_{\mathrm{gzy}}& k_{\mathrm{gzz}}\end{array}\right)\&CenterDot;\left(\begin{array}{c}{\mathrm{\&omega;}}_x\\ {\mathrm{\&omega;}}_y\\ {\mathrm{\&omega;}}_z\end{array}\right)+\left(\begin{array}{c}{B}_{\mathrm{gx}}\\ B_{\mathrm{gy}}\\ B_{\mathrm{gz}}\end{array}\right)...\left(28\right)$

F wherein _Ai, f _GiBe respectively i axis accelerometer, gyro output, A _i, ω _iBe respectively i axis accelerometer, gyro input, B _Ai, B _GiBe respectively i axis accelerometer, gyro zero partially, k _Aii, k _GiiBe respectively input i axis accelerometer, the projection of gyro unit's input on output i axle, k _Aij, k _GijBe respectively projection on output i axle of input j axis accelerometer, the input of gyro unit (i, j=x, y, z).

Step 2: inertial navigation system is installed on the double axle table according to three modes of pointing to the sky, northeast, collecting device is coupled together through turntable connecting line and inertial navigation system, the inertial navigation system output data is transferred to collecting device through serial communication mode.Utilize stabilized voltage supply that inertial navigation system is supplied power, energising surpassed after 30 minutes, saw Fig. 4, got into step 3.

Step 3: according to 12 positions shown in Figure 1 be sky, northeast, southwest sky, east northeast ground, Nan Xidi, day east southeast, sky northwest (NW), east northeast, Di Nanxi, Nan Tian west, Bei Tiandong, Nan Didong, western 12 positions, backlands; Static state was gathered three minutes on each position respectively; Output data is shown through display interface; With the observed data correctness, and image data got average and store the demarcation matrix into.

Step 4: according to sky, northeast, two positions, northwest (NW) sky shown in Figure 2, be ± 2 °/s around z, y, x with ω respectively, ± 10 °/s; ± 30 °/s; ± 60 °/s, ± 100 °/s rotated three minutes, and output data is shown through display interface; With the observed data correctness, and image data got average and store the demarcation matrix into.

Step 5: according to four positions shown in the accompanying drawing 3, the static collection 15 minutes on each position shows output data through display interface respectively, with the observed data correctness, and image data got average and stores the demarcation matrix into.

Step 6:

1) utilize least square method to resolve the accelerometer calibrating parameters:

Utilize the measurement result of 12 positions in the step 3, wherein

expression x axle is towards the mean value of all position j axle outputs in sky;

expression x axle is towards the mean value of all position j axle outputs on ground;

expression y axle is towards the mean value of all position j axle outputs in sky;

expression y axle is towards the mean value of all position j axle outputs on ground;

expression z axle is towards the mean value of all position j axle outputs in sky;

expression z axle is towards the mean value of all position j axle outputs on ground.

$B_{\mathrm{ax}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z-})/6...\left(29\right)$

$k_{\mathrm{axx}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x-})/2g...\left(30\right)$

$k_{\mathrm{axy}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y-})/2g...\left(31\right)$

$k_{\mathrm{axz}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z-})/2g...\left(32\right)$

$B_{\mathrm{ay}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z-})/6...\left(33\right)$

$k_{\mathrm{ayx}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x-})/2g...\left(34\right)$

$k_{\mathrm{ayy}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y-})/2g...\left(35\right)$

$k_{\mathrm{ayz}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z-})/2g...\left(36\right)$

$B_{\mathrm{az}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z-})/6...\left(37\right)$

$k_{\mathrm{azx}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x-})/2g...\left(38\right)$

$k_{\mathrm{azy}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y-})/2g...\left(39\right)$

$k_{\mathrm{azz}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z-})/2g...\left(40\right)$

2) utilize analytical method to resolve gyro constant multiplier and misalignment:

Utilize three rotating measurement results in the step 4, wherein

For (j=x, y is z) with angular speed ± ω around j _kBe slewing rate, in rotation n time-of-week, (z) axle is respectively at angular speed ω for i=x, y for i _k,-ω _kMean value poor of output, ω ₁, ω ₂, ω ₃, ω ₄, ω ₅Be respectively 2 °/s, 10 °/s, 30 °/s, 60 °/s, 100 °/s.

$k_{\mathrm{gx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(41\right)$

$k_{\mathrm{gy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(42\right)$

$k_{\mathrm{gz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png,HDA0000139326040000012.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="3"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="4"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;<figure-callout\; id="5"\; label="F\; gzz"\; filenames="HDA0000139326040000011.png"\; state="\{\{state\}\}">F</figure-callout>}}_{\mathrm{gzz}}^{}}}{{2\mathrm{\&omega;}}_5})/5...\left(43\right)$

$E_{\mathrm{gxy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(44\right)$

$E_{\mathrm{gxz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(45\right)$

$E_{\mathrm{gyx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(46\right)$

$E_{\mathrm{gyz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(47\right)$

$E_{\mathrm{gzx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(48\right)$

$E_{\mathrm{gzy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(49\right)$

3) utilize analytical method to resolve gyro zero partially

B _gx＝(f _1x+f _2x+f _3x+f _4x)/4............................................(50)

B _gy＝(f _1y+f _2y+f _3y+f _4y)/4............................................(51)

B _gz＝(f _1z+f _2z+f _3z+f _4z)/4............................................(52)

Utilize measurement result in the step 5, f _IjRepresent i the position output of j axle gyro (i=1,2,3,4; J=x, y, z).

Step 7: simulated conditions: local latitude is set to 39.9778 °, and acceleration of gravity is 9.80158m/s ², the normal value zero of gyro combination be respectively partially 0.01 °/h add Gauss's white noise (standard variance be 0.005 °/h), the normal value of accelerometer combination is zero to be that 50 μ g add Gauss's white noise (standard variance is 10 μ g) partially.Turntable install lateral error be respectively 3 ', the north orientation error is 1 °.

1) accelerometer emulation:

Emulation contains the inertial navigation system output data of turntable alignment error, carries out six positions, 12 location positions.Static six positions, 12 position measurements 3 minutes.Calibration result (1) expression six location position results, calibration result (2) expression 12 location position results.Simulation result is as shown in table 1:

Table 1 accelerometer is demarcated simulation result

Can see that by calibration result (1) it is bigger that calibrating parameters is influenced by the turntable alignment error, x axle, z axle zero be error partially, k _Gyz, k _GzyReached the true value magnitude, other each item calibrated errors have also exceeded navigation level inertial navigation system calibration request.Can be known that by calibration result (2) the turntable alignment error is not demarcated accelerometer bias and exerted an influence, be consistent to the ratio that influences of misalignment influence, constant multiplier.Under the turntable alignment error condition of setting, the error amount of constant multiplier is 0.8ppm, and it is less to influence magnitude.Above result shows that 12 position accelerometer scaling methods can effectively eliminate turntable alignment error influence, obtains accurate calibration result.

2) Gyro Calibration emulation

Emulation contains the inertial navigation system output data of turntable alignment error, utilizes aforementioned Gyro Calibration method fixed through rower, with 5 of three rotatings of 10 °/s, and static 4 position measurements 3 minutes.Calibration result representes to contain the calibration result of turntable error.Simulation result is as shown in table 2:

Table 2 Gyro Calibration simulation result

Can know that through simulation result calibration result is consistent with the emulation true value, show that the design can accurately separate calibrating parameters, obtains calibration result.

Claims (1)

1. scaling method based on the optics SINS of double axle table, it is characterized in that: the concrete cloth of this method is following suddenly:

Step 1: the choosing of error model;

The accelerometer error model:

$\left(\begin{array}{c}{f}_{\mathrm{ax}}\\ f_{\mathrm{ay}}\\ f_{\mathrm{az}}\end{array}\right)=\left(\begin{array}{ccc}{k}_{\mathrm{axx}}& k_{\mathrm{axy}}& k_{\mathrm{axz}}\\ k_{\mathrm{ayx}}& k_{\mathrm{ayy}}& k_{\mathrm{ayz}}\\ k_{\mathrm{azx}}& k_{\mathrm{azy}}& k_{\mathrm{azz}}\end{array}\right)\&CenterDot;\left(\begin{array}{c}{A}_x\\ A_y\\ A_z\end{array}\right)+\left(\begin{array}{c}{B}_{\mathrm{ax}}\\ B_{\mathrm{ay}}\\ B_{\mathrm{az}}\end{array}\right)...\left(1\right)$

Gyroscope error model:

$\left(\begin{array}{c}{f}_{\mathrm{gx}}\\ f_{\mathrm{gy}}\\ f_{\mathrm{gz}}\end{array}\right)=\left(\begin{array}{ccc}{k}_{\mathrm{gxx}}& k_{\mathrm{gxy}}& k_{\mathrm{gxz}}\\ k_{\mathrm{gyx}}& k_{\mathrm{gyy}}& k_{\mathrm{gyz}}\\ k_{\mathrm{gzx}}& k_{\mathrm{gzy}}& k_{\mathrm{gzz}}\end{array}\right)\&CenterDot;\left(\begin{array}{c}{\mathrm{\&omega;}}_x\\ {\mathrm{\&omega;}}_y\\ {\mathrm{\&omega;}}_z\end{array}\right)+\left(\begin{array}{c}{B}_{\mathrm{gx}}\\ B_{\mathrm{gy}}\\ B_{\mathrm{gz}}\end{array}\right)...\left(2\right)$

F wherein _Ai, f _GiBe respectively i axis accelerometer, gyro output, A _i, ω _iBe respectively i axis accelerometer, gyro input, B _Ai, B _GiBe respectively i axis accelerometer, gyro zero partially, k _Aii, k _GiiBe respectively input i axis accelerometer, the projection of gyro unit's input on output i axle, k _Aij, k _GijBe respectively projection on output i axle of input j axis accelerometer, the input of gyro unit (i, j=x, y, z);

Step 2: direction is installed on the double axle table towards the sky, northeast with three of inertial navigation systems, and turntable, inertial navigation system, collecting computer connect and finish, and the energising preheating got into data acquisition after 30 minutes;

Step 3: according to 12 positions be sky, northeast, southwest sky, east northeast ground, Nan Xidi, day east southeast, sky northwest (NW), east northeast, Di Nanxi, Nan Tian west, Bei Tiandong, Nan Didong, western position, backlands; Each position is static gathered three minutes; And after making even data all, store into and demarcate in the matrix;

Step 4: according to two positions is sky, northeast, position, northwest (NW) sky, is ± 2 °/s around z, y, x with ω respectively, and ± 10 °/s, ± 30 °/s, ± 60 °/s, ± 100 °/s rotated three minutes, stores into after making even output data all and demarcates in the matrix;

Step 5: according to four positions is sky, northeast, southwest sky, east northeast ground, southern western destination location, and static collection 15 minutes on each position is respectively got output data average and stored in the demarcation matrix;

Step 6: utilize least square method to resolve the accelerometer calibrating parameters, utilize analytic calculation Gyro Calibration parameter;

1) utilize least square method to resolve the accelerometer calibrating parameters:

Utilize the measurement result of 12 positions in the step 3, wherein

expression x axle is towards the mean value of all position j axle outputs in sky;

expression x axle is towards the mean value of all position j axle outputs on ground;

expression y axle is towards the mean value of all position j axle outputs in sky;

expression y axle is towards the mean value of all position j axle outputs on ground;

expression z axle is towards the mean value of all position j axle outputs in sky; expression z axle is towards the mean value of all position j axle outputs on ground;

$B_{\mathrm{ax}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z-})/6...\left(3\right)$

$k_{\mathrm{axx}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{x-})/2g...\left(4\right)$

$k_{\mathrm{axy}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{y-})/2g...\left(5\right)$

$k_{\mathrm{axz}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ax}}^{z-})/2g...\left(6\right)$

$B_{\mathrm{ay}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z-})/6...\left(7\right)$

$k_{\mathrm{ayx}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{x-})/2g...\left(8\right)$

$k_{\mathrm{ayy}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{y-})/2g...\left(9\right)$

$k_{\mathrm{ayz}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{ay}}^{z-})/2g...\left(10\right)$

$B_{\mathrm{az}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y-}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z+}+{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z-})/6...\left(11\right)$

$k_{\mathrm{azx}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{x-})/2g...\left(12\right)$

$k_{\mathrm{azy}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{y-})/2g...\left(13\right)$

$k_{\mathrm{azz}}=({\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z+}-{\stackrel{\&OverBar;}{f}}_{\mathrm{az}}^{z-})/2g...\left(14\right)$

2) utilize analytical method to resolve gyro constant multiplier and misalignment:

Utilize three rotating measurement results in the step 4, wherein

For (j=x, y is z) with angular speed ± ω around j _kBe slewing rate, in rotation n time-of-week, (z) axle is respectively at angular speed ω for i=x, y for i _k,-ω _kMean value poor of output, ω ₁, ω ₂, ω ₃, ω ₄, ω ₅Be respectively 2 °/s, 10 °/s, 30 °/s, 60 °/s, 100 °/s;

$k_{\mathrm{gx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(15\right)$

$k_{\mathrm{gy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(16\right)$

$k_{\mathrm{gz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzz}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzz}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzz}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzz}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzz}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(17\right)$

$E_{\mathrm{gxy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(18\right)$

$E_{\mathrm{gxz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gxz}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(19\right)$

$E_{\mathrm{gyx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(20\right)$

$E_{\mathrm{gyz}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gyz}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(21\right)$

$E_{\mathrm{gzx}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^2}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzx}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(22\right)$

$E_{\mathrm{gzy}}=(\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^1}}{{2\mathrm{\&omega;}}_1}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}}}{{2\mathrm{\&omega;}}_2}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^3}}{{2\mathrm{\&omega;}}_3}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^4}}{{2\mathrm{\&omega;}}_4}+\frac{\stackrel{\&OverBar;}{{\mathrm{\&Delta;F}}_{\mathrm{gzy}}^5}}{{2\mathrm{\&omega;}}_5})/5...\left(23\right)$

3) utilize analytical method to resolve gyro zero partially

B _gx＝(f _1x+f _2x+f _3x+f _4x)/4..............................................(24)

B _gy＝(f _1y+f _2y+f _3y+f _4y)/4..............................................(25)

B _gz＝(f _1z+f _2z+f _3z+f _4z)/4..............................................(26)

Utilize measurement result in the step 5, f _IjExpression i the position output of j axle gyro (i=1,2,3,4; J=x, y, z).

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